[最近一次求职面试中,我要求给出第三类斐波那契数列的第100个元素的结果(Fib(n)= Fib(n-1)+ Fib(n-2)+ Fib(n-3)我完成了数学归纳法,并构造了一个表示大于long long的数字的类,然后要求我通过模板元编程来实现它,问题是结果将超出long long的范围,我不知道如何解决此问题。这是我使用模板元编程的代码。
template<long long num>
struct fib
{
enum { result = fib<num - 1>::result + fib<num - 2>::result + fib<num - 3>::result};
};
template<>
struct fib<0>
{
enum { result = 1 };
};
template<>
struct fib<1>
{
enum { result = 1 };
};
template<>
struct fib<2>
{
enum { result = 2 };
};
template<>
struct fib<3>
{
enum { result = 4 };
};
int main()
{
cout << fib<100>::result << endl;
return 0;
}
一种可能的实现是使用自定义结构来存储数字,而不是内置类型。例如,您可以存储这样的数字:
template <int... Digits>
struct number<Digits... > { };
为了简单起见,在添加时,我以相反的顺序存储数字,因此数字275
被存储为number<5, 7, 2>
。
Fibonacci只需要加法,因此您只需定义加法,例如模板add
(有关实际实现,请参见答案的结尾)。
然后您可以非常轻松地定义fib
模板:
template <int N>
struct fib_impl {
using type = add_t<
typename fib_impl<N-1>::type,
typename fib_impl<N-2>::type,
typename fib_impl<N-3>::type>;
};
template <>
struct fib_impl<0> { using type = number<0>; };
template <>
struct fib_impl<1> { using type = number<0>; };
template <>
struct fib_impl<2> { using type = number<1>; };
template <int N>
using fib = typename fib_impl<N>::type;
并且使用适当的输出运算符(请参见下文),您可以打印第100个Tribonacci数字:
int main() {
std::cout << fib<100>{} << "\n";
}
operator<<
的实现:std::ostream& operator<<(std::ostream &out, number<>) {
return out;
}
template <int Digit, int... Digits>
std::ostream& operator<<(std::ostream &out, number<Digit, Digits... >) {
// Do not forget that number<> is in reverse order:
return out << number<Digits... >{} << Digit;
}
add
模板的实现:cat
实用程序:// Small concatenation utility:
template <class N1, class N2>
struct cat;
template <int... N1, int... N2>
struct cat<number<N1... >, number<N2... >> {
using type = number<N1... , N2...>;
};
template <class N1, class N2>
using cat_t = typename cat<N1, N2>::type;
template <class AccNumber, int Carry, class Number1, class Number2>
struct add_impl;
template <class AccNumber, int Carry>
struct add_impl<AccNumber, Carry, number<>, number<>> {
using type = std::conditional_t<Carry == 0, AccNumber, cat_t<AccNumber, number<1>>>;
};
template <class AccNumber, int Carry,
int Digit2, int... Digits2>
struct add_impl<AccNumber, Carry, number<>, number<Digit2, Digits2...>> {
using type = typename add_impl<
cat_t<AccNumber, number<(Digit2 + Carry) % 10>>,
(Digit2 + Carry) / 10,
number<Digits2... >, number<>>::type;
};
template <class AccNumber, int Carry,
int Digit1, int... Digits1>
struct add_impl<AccNumber, Carry, number<Digit1, Digits1... >, number<>> {
using type = typename add_impl<
cat_t<AccNumber, number<(Digit1 + Carry) % 10>>,
(Digit1 + Carry) / 10,
number<Digits1... >, number<>>::type;
};
template <class AccNumber, int Carry,
int Digit1, int... Digits1, int Digit2, int... Digits2>
struct add_impl<AccNumber, Carry, number<Digit1, Digits1... >, number<Digit2, Digits2...>> {
using type = typename add_impl<
cat_t<AccNumber, number<(Digit1 + Digit2 + Carry) % 10>>,
(Digit1 + Digit2 + Carry) / 10,
number<Digits1... >, number<Digits2... >>::type;
};
template <class... Numbers>
struct add;
template <class Number>
struct add<Number> {
using type = Number;
};
template <class Number, class... Numbers>
struct add<Number, Numbers... > {
using type = typename add_impl<
number<>, 0, Number, typename add<Numbers... >::type>::type;
};
template <class... Numbers>
using add_t = typename add<Numbers... >::type;
我不知道模板的现成的速动精度功能。但是,玩具数字类型很容易写:
template <long long H,long long L>
struct my_number {
static const long long high = H;
static const long long low = L;
static const long long mod = 10000000000;
static void print() {
std::cout << high << setw(10) << setfill('0') << low;
}
};
将结果的最后10
位存储在low
中,将前导位存储在high
中。可以通过
my_number
相加template <typename A,typename B>
struct sum {
static const long long low = (A::low + B::low) % A::mod;
static const long long high = A::high + B::high + (A::low + B::low) / A::mod;
using number = my_number<high,low>;
};
以及3个数字:
template <typename A,typename B,typename C>
struct sum3 { using number = typename sum<A,sum<B,C>>::number; };
如上所述,这只是一个玩具示例。无论如何,一旦您的数字类型可以表示足够大的数字,则只需稍加修改即可调整fib
:
template<long long num> struct fib {
using result_t = typename sum3< typename fib<num-1>::result_t,
typename fib<num-2>::result_t,
typename fib<num-3>::result_t
>::number;
};
template<> struct fib<0> { using result_t = my_number<0,1>; };
template<> struct fib<1> { using result_t = my_number<0,1>; };
template<> struct fib<2> { using result_t = my_number<0,2>; };
template<> struct fib<3> { using result_t = my_number<0,4>; };
int main() {
fib<100>::result_t::print();
}
我找不到正确值fib<100>
的可靠来源,因此很遗憾,我无法对此进行测试。完整示例为here。